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Let $S$ be a submanifold of a real smooth manifold $M$. By a splitting of the normal bundle of $S$ I mean a sub-bundle $V$ of $TM|_S$ such that $V\oplus TS = TM|_S$.

Question: Given such a splitting, can I always find local coordinates $x_1,\dots,x_s,y_1,\ldots, y_r$ on $M$ around each point $p\in S$, such that the submanifold $S$ is given by $y_1=0,\ldots, y_r=0$ and $V$ is given by the kernels of $dx_1,\ldots,dx_s$?

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Yes, one can always do this. Start with a $p$-centered local coordinate system $(x^\sigma,y^\rho)$ on an open $p$-neighborhood $U\subset M$ such that $S\cap U$ is given by $y^\rho=0$ $(1\le\rho\le r)$. Then there will exist functions $F^\sigma_\rho(x)$ such that $V$ along $S\cap U$ is given by the equations $$ \mathrm{d}x^\sigma + F^\sigma_\rho(x)\,\mathrm{d}y^\rho = 0 $$ (summation convention assumed). This is the same as $$ \mathrm{d}(x^\sigma+F^\sigma_\rho(x)y^\rho) - y^\rho\,\mathrm{d}(F^\sigma_\rho(x)) = 0. $$ Since $y^\rho=0$ along $S$, this says that $V$ along $S$ is defined by $$ \mathrm{d}(x^\sigma+F^\sigma_\rho(x)y^\rho) = 0, $$ so take $\bar x^\sigma = x^\sigma+F^\sigma_\rho(x)y^\rho$, and let $(\bar x^\sigma,y^\rho)$ be your new coordinate system. (One can easily check that these are local coordinates in an open neighborhood of $S\cap U\subset U$.)

Added Remark: Just a comment on your terminology: Literally, what you are defining is not a splitting of the normal bundle of $S$, but rather a choice of a normal bundle along $S$, and you are asking whether there are any local invariants that distinguish such choices. As the above shows, the answer is 'no'. In fact, there is even a global version: For two normal bundles $V_1$ and $V_2$ along a closed submanifold $S$, there is a diffeomorphism of $M$ that fixes $S$ and carries $V_1$ to $V_2$.

Note also that, in the holomorphic category (in contrast to the smooth case), normal bundles need not be unique in this sense, and may not even exist.

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  • $\begingroup$ Sorry for my sloppy use of terminology. I had considered using "choice of normal bundle", but in many places the quotient $TM/TS$ is called the normal bundle, and there's no choice involved. I ended up talking about the splitting of $TM/TS$, but what I meant was the splitting of the short exact sequence. $\endgroup$ Commented Aug 29, 2017 at 12:26
  • $\begingroup$ @MichaelBächtold: Of course, 'normal' is already a misleading descriptor for this concept, since there is no inner product involved. A better term might have been a 'transversal subbundle' or a 'complementary subbundle', while TM/TS could have been called the 'tangential quotient bundle' or some such. However, we can't fix all of these misleading terminologies. $\endgroup$ Commented Aug 29, 2017 at 13:11
  • $\begingroup$ The only natural definition of the normal bundle to $S$ in $M$ is, just as Michael writes, $N=\frac {TM\vert S}{TS}$. This is the definition adopted by all algebraic geometers and complex analysts. There is NO natural way to see that bundle $N$ as a subbundle of $TM\vert S$: don't we strictly forbid our students to confuse a quotient vector space with a subvector space? (To be continued) . $\endgroup$ Commented Feb 1, 2019 at 17:18
  • $\begingroup$ Of course given some supplementary structure on a manifold such as,say, a Riemannian metric we can identify the quotient bundle with a subbundle , but conceptually it is better to consider the normal bundle as a quotient of the restricted tangent bundle. $\endgroup$ Commented Feb 1, 2019 at 17:25

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